# What does E=mc^2 mean?

1. Mar 26, 2009

### quZz

Could someone explain to me, what does $$E=mc^2$$ mean? I've got a feeling that this famous relation is just a definition of m.

First, if we use an old version
$$E=mc^2\:\:(1)$$
where E is total energy and m is relativistic mass, then is there a way to measure m and E independently so that this equation becomes a physical law?
In fact, if we are dealing with a free moving particle than $$\textbf{p} = E\textbf{v}/c^2$$. Now if we want to folow Newton and define $$\textbf{p} = m\textbf{v}$$ we get (1).
In particle interactions it is 4-momentum $$p = (E/c,\:\textbf{p})$$ that is conserved.
If we try to weigh smth then (as it follows from GR) it is $$T_{\mu \nu}$$ that matters. Actually the gravitational force acting on a moving particle can be different depending on it's moving direction. In the simplest case only $$T_{00}$$ survive, but again it is energy density.

Second, if we use a modern one
$$E_0=mc^2\:\:(2)$$
with E being rest energy, then it follows right from the definition of 4-momentum norm
$$mc \equiv \sqrt{(E/c)^2-\textbf{p}^2}$$.
If we recall Einstein's derivation of (2), it was based on the fact that c.m. of a closed system is not accelerating, but in SR c.m. is defined... again from energies of particles.

So please explain to me, what is physical about (1) or (2) or both?

2. Mar 26, 2009

### HallsofIvy

Actually, no. "m" in that equation is NOT the relativistic mass. It is the rest mass. It couldn't be the relativistic mass (which many physicists no longer use) because that depends on the frame of reference.

3. Mar 26, 2009

### quZz

Of course in (2) I meant "m" to be the rest mass, sorry if that was not clear. The question still holds though =) is there a way E_0 and m can be defined independently?

4. Mar 26, 2009

### clem

For most physicists, m, is the "invariant mass", defined by the Lorentz scalar
m^2=(E^2-p^2c^2)/c^4. In the rest system, defined by p=0, m=E_0/c^2,
where E_0 is the 0 component of the 4-vector (E,pc) when p=0.
That is, m and E are different objects, one being a scalar and the other the 0 component of a 4-vector. They are related by the equations above.
This is like saying that the length L of a rod equals L_x if L_y and L_z are zero.

5. Mar 26, 2009

### quZz

And what's wrong with that? To measure the length you just place a ruler (X axis) parallel to the rod so that L_y = L_z = 0 and then |L_x| is equal to L by definition, am I wrong?

6. Mar 26, 2009

### clem

I guess I wasn't clear enough. L and L_x are equal in magnitude when L_y and L_z are zero. But |L_x| is not equal to L "by definition". By definition,
$$L=\sqrt{L_x^2+L_y^2+L_z^2}$$.
That was my example of a simple case where two things are equal in one system, but are different and not equal in a general system. That is, if you rotate coordinates L and L_x are no longer equal.

7. Mar 26, 2009

### Naty1

The equation E = mc2 simply reflects the fact that a small amount of mass can be used as the source for a lot of energy. Or as Einstein originally looked at it,
E/c2 = m....reflecting that takes a lot of energy to make a little mass....

It does not explain why these are so, it only explains the quantative relationship.... it does not explain energy nor mass nor the speed of light....in fact we do not know what mass, for example, even is...but we can describe a lot of its attributes, its behaviors, from such predictive formulas.....

8. Mar 26, 2009

### quZz

Clem, that is quite right. But $$E$$ is not equal to $$mc^2$$ =) $$E_0$$ is. It is measured always with respect to one fixed observer. So if you rotate your 4-coordinate system properly you'll get $$p = (E_0/c, \textbf{0})$$ and $$E_0 = m c^2$$ just as if you rotate a ruler over the rod so that $$L_y = L_z = 0$$ and $$L_x = L$$.
Really, if you have rotated the ruler (4-coordinate system) properly then what is the difference between $$L_x$$ and $$L$$ ($$E_0/c$$ and $$m c$$)? I think there is no difference at all and the former is the definition of the latter.

I want to make myself clear. Consider a magnetic field of stationary moving charges. Using magnetometer you can measure magnetic field $$\textbf{H}$$ in every point in space. You can measure charge velocity $$\textbf{v}$$ and charge density $$\rho$$ (by it's electric field). So all these quantities can be measured independently. But it turns out that Ampere's law always holds $$\nabla \times \textbf{H} = \rho \textbf{v}/c$$. That's what I'm calling a physical law. It seems to me that $$E_0 = mc^2$$ is not a physical law in that sense because $$E_0$$ and $$mc$$ are just two different labels of the same thing.
Am I wrong? =)

9. Mar 26, 2009

### lightarrow

You have a piece of black paper and you weigh it: 1g. Then you put it under the sun, so that it absorbs light energy, and while still hot, you weigh it again: 1.00...1 g. Could you have predicted this result without the knowledge that E_0 = mc^2?

10. Mar 27, 2009

### quZz

Yes, since in general relativity it is energy (in general case all components of $$T_{\mu\nu}$$) that is responsible for gravitational interaction. It would be even simpler not to bother about mass at all... just add the energy absorbed from the sun and you get additional force that you call the weight of the piece of paper.

11. Mar 27, 2009

### lightarrow

Ah, and you can have general relativity without knowing E_0 = mc^2 which is a relativistic result? Weird.

12. Mar 27, 2009

### quZz

Derivation of Einstein's equations has nothing to do with E_0 = mc^2 =))

13. Mar 27, 2009

### lightarrow

Do you want to start from Einstein's equations without known anything about relativity? Everything is connected, isnt'it? E_0 = mc^2 is connected with relativity which is connected with Einstein's equations; you can't take one and not the other.
Anyway, if you mean that you could only speak of "invariant energy/c^2" instead of "mass", then I agree with you; however "mass" is shorter .

14. Mar 27, 2009

### quZz

that was a bit too philozophical for me
Something like that... I want to say that there is no mass as an independent quantity in relativity any more. So this law is not physical law (in the sense i mentioned above) but rather a clarification that what was previously known as mass in fact appears to be another form of energy.

15. Mar 27, 2009

### Ben Niehoff

Energy in physics is ultimately just a cost function: it represents how much it "costs" for any given sequence of events to happen. The relation E=mc^2, then, simply represents the cost of creating something of mass m; or conversely, it represents the available ability to do work if a quantity of mass m were destroyed (and converted to work).

16. Mar 27, 2009

### intrepid_nerd

E $$\neq$$ mc2
E = (mc2) / $$\sqrt{}$$1-q2/c2)

q = velocity [q is what Einstein originally used]

the denominator is meant to account for, obviously, velocity, it's a calibration. so, as an object's velocity reaches the speed of light, the root of the calibration decreases and E increases, as far as we think this is from Newtonian, it's all there. to accelerate, a force is needed. BUT to combine with Maxwell, there's also the option of sacrificing mass for energy to accelerate, thus keeping constant E yet obtaining acceleration (is this right? i've never thought about it). Think about a person being pushed on a swing, there is no caloric transformation going on in their body to create the energy required to swing, as someone would if they were not being pushed. This equation provides a bridge between Newton and Maxwell.

THIS, TO ME, BRINGS ABOUT A HUGE CONUNDRUM!!
take three different time intervals of an accelerating particle, perhaps let q = at.
t0 we'll have be a time when q << c -- the equation works flawlessly, we see that the equation accounts for GR and there is no theoretical debate.
t2 we'll have be a time when q < c (but just barely) -- the particle has accelerated and q is very close to c, meaning that either the mass has decreased substantially or there has been a major increase in energy.
t2 we'll let be when q = c -- then either m$$\rightarrow$$0 BEFORE q = c, thus q will never actually reach c and the system dissipates, or E$$\rightarrow\infty$$...buuut what does that mean? does the universe implode? does the particle become everything and nothing? does it cross the space/time fabric to absorb energy from elsewhere?

my conundrum is that photons are massless (yet have momentum), quantized packets of energy moving at c. take a photon with an arbitrary amount of energy. then using above equation we get 0/0? i'm lost.

ending with a question: can anyone tell me why Einstein took the root of 1 - q2/c2? could you perhaps offer some sort of way to picture it?

17. Mar 27, 2009

### intrepid_nerd

somewhere in quantum chromodynamics they're finding that gluons are the source of mass...at least that's what i got out of an article last month, can't say i understood much in it.

18. Mar 27, 2009

### quZz

Dear intrepid_nerd, I think you should read something on SR first =) all your CONUNDRUM!!!s will vanish rapidly...

19. Mar 28, 2009

### intrepid_nerd

SR? string... relativity? do you mean GR?
just started this week reading about GR so don't know anything about it, chemistry is what i know.

20. Mar 28, 2009

### quZz

I mean Special Relativity